If n is a term of this sequence, then n^2 = a^3 + b^3 = c^3 + d^3 where (a, b) and (c, d) are distinct pairs. If n^2 = a^3 + b^3 = c^3 + d^3, then (n*k^3)^2 = n^2*k^6 = k^6*(a^3 + b^3) = k^6*(c^3 + d^3) = (a*k^2)^3 + (b*k^2)^3 = (c*k^2)^3 + (d*k^2)^3. It is obvious that if (a, b) and (c, d) are distinct, then (k^2*a, k^2*b), (k^2*c, k^2*d) are also distinct for all nonzero values of k. So if n is in this sequence and n*k^3 is not in A155961, then n*k^3 is in this sequence for all k > 0. If this sequence is not infinite, then there are infinitely many consecutive k values for any term n such that n*k^3 is in A155961. Is it possible? - Altug Alkan, May 10 2016